A charged particle of mass m = 2 kg and char | vedclass

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(A) The horizontal acceleration due to the electric field is $a_x = \frac{qE}{m} = \frac{10^{-6} 	imes 2 	imes 10^7}{2} = 10 \ ms^{-2}$.The time of

Vedclass · Accepted Answer

$20$

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(A) The horizontal acceleration due to the electric field is $a_x = \frac{qE}{m} = \frac{10^{-6} 	imes 2 	imes 10^7}{2} = 10 \ ms^{-2}$.The time of flight $T$ is determined by the vertical motion: $T = \frac{2u \sin 	heta}{g} = \frac{2 	imes 10 	imes \sin 45^{\circ}}{10} = \frac{2 	imes 10 	imes \frac{1}{\sqrt{2}}}{10} = \sqrt{2} \ s$.The horizontal range $R$ is given by the kinematic equation $R = u_x T + \frac{1}{2} a_x T^2$.Here,$u_x = u \cos 45^{\circ} = 10 	imes \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \ ms^{-1}$.Substituting the values: $R = (\frac{10}{\sqrt{2}}) 	imes \sqrt{2} + \frac{1}{2} 	imes 10 	imes (\sqrt{2})^2$.$R = 10 + \frac{1}{2} 	imes 10 	imes 2 = 10 + 10 = 20 \ m$.

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